Let $(X_N)_N$ be a sequence of trace class operators acting on, say, $L^2(\mathbb{R})$. What are the minimal assumptions in order to have the convergence of their Fredholm determinant
$$
\lim_N\det(I+X_N) ?
$$
I know $X\mapsto \det(I+X_N)$ is continuous for the trace class norm topology (once restricted to trace class), ok. Let's say that $X_N$ converges weakly to $X$, it is enough to have the convergence of Fredholm determinants ? What conditions should we add ? I guess I just need a good reference on the topic. Any ideas ? Thanks in advance.

1 Answer
1

The Fredholm determinant is not sequentially continuous in the strong topology. Take the sequence of 1-dimensional projectors $X_n=\langle e_n,\cdot\rangle e_n$ with the $e_n$ forming an ON basis. You then have $X_n \to 0$ strongly, but $0=\det(I-X_n)$ does not converge to $1=\det(I)$. Or what did you have in mind when saying that you would know the continuity for the strong topology (once restricted to trace class)?

If you, however, add the convergence $\|X_n\|_1\to\|X\|_1$, where $\|\cdot\|_1$ denotes trace class norm, then it is known that weak sequential convergence $X_n \to X$ implies convergence in trace class norm and, therefore, of the Fredholm determinant (see Thm. 2.21 and Addendum H of the book "Trace Ideals and Their Applications" by Barry Simon, 2nd ed., AMS 2005).

Folkmar, do you know other results if one restrict to operators which are symmetric positive integral kernel operators ? Anyway, thx for your answer
–
Adrien HardyAug 12 '11 at 12:07

Weak operator convergence and convergence of the trace (which is then, in the case of symmetric positive integral trace class operators, just the trace class norm) will do. Have a look at Chapter 2 of Simon's book.
–
Folkmar BornemannAug 12 '11 at 15:59